Calculate the pH of a 0.35 M solution of dimethylamine, (CH3)2NH.
To calculate the pH of a solution of dimethylamine, we need to consider its properties as a weak base and use the equation for the ionization of the base in water.
Dimethylamine, (CH3)2NH, is a weak base that partially ionizes in water according to the equation:
(CH3)2NH + H2O ⇌ (CH3)2NH2+ + OH-
In this equation, (CH3)2NH represents the dimethylamine molecule, (CH3)2NH2+ represents the conjugate acid (formed when dimethylamine accepts a proton), and OH- represents the hydroxide ion produced by the ionization of water.
Now, let's assume that x represents the concentration of (CH3)2NH that ionizes, and since dimethylamine is a weak base, we can assume that the concentration of (CH3)2NH2+ and OH- are equal to x.
Then, we can write an expression for the equilibrium constant, Kb, for the ionization of dimethylamine:
Kb = ( [ (CH3)2NH2+ ] * [ OH- ] ) / [ (CH3)2NH ]
Now, since we know the initial concentration of dimethylamine is 0.35 M, and at equilibrium, the concentration of (CH3)2NH is (0.35 - x), we can rewrite the expression for Kb as:
Kb = ( x^2 ) / (0.35 - x)
To proceed, we need to know the value of Kb for dimethylamine, which is 5.4 x 10^-4 M at 25°C. Using this value, we can solve for x by rearranging the equation to:
(5.4 x 10^-4) = ( x^2 ) / (0.35 - x)
This is a quadratic equation. By solving it, we find that x ≈ 0.0188 M.
Since x represents the concentration of OH- ions, we know that the concentration of OH- ions is 0.0188 M.
To calculate the pOH, we can use the equation:
pOH = -log10([OH-])
pOH = -log10(0.0188) ≈ 1.73
Finally, to calculate the pH, we use the equation:
pH + pOH = 14
pH + 1.73 ≈ 14
pH ≈ 14 - 1.73 ≈ 12.27
Therefore, the pH of a 0.35 M solution of dimethylamine is approximately 12.27.